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\newtheorem{definition}{Definition}[section]%定义
\newtheorem{theorem}{Theorem}[section]%定理
\newtheorem{axiom}{Axiom}[section]%公理
\newtheorem{lemma}{Lemma}[section]%引理
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\title{\heiti\zihao{2} 习题16.4}
\author{中书君}
\date{\today}
\begin{document}
\maketitle
\section{求曲面 $z=x y$ 包含在圆柱面 $x^{2}+y^{2}=1$ 内部分曲面的面积.}
\textbf{解}\quad
\begin{tikzpicture}
    \begin{axis}
        \addplot3[mesh]{x*y};
    \end{axis}
\end{tikzpicture}
$$
\begin{aligned}
    S=\iint\limits_D\sqrt{1+\left(\dfrac{\partial z}{\partial x}\right)^2+\left(\dfrac{\partial z}{\partial y}\right)^2}\mathrm{d}x\mathrm{d}y&=\int_0^{2\pi}\mathrm{d}\theta\int_0^1\rho\sqrt{1+\rho^2}\mathrm{d}\rho\\
    &=\dfrac{2\pi}{3}(2\sqrt{2}-1)
\end{aligned}
$$

\section{求圆锥面 $x^{2}+y^{2}=\dfrac{1}{3} z^{2}(z \geqslant 0)$ 被平面 $x+y+z=2$ 所截部分曲面的面积.}
\textbf{解}\quad
由于圆锥面各处的法向量与$z$轴夹角相同,从而有
$$
\begin{aligned}
    \iint\limits_D\dfrac{1}{|\cos<\boldsymbol{n},z>|}\mathrm{d}x\mathrm{d}y&=\dfrac{1}{|\cos<\boldsymbol{n},z>|}\iint\limits_D\mathrm{d}x\mathrm{d}y\\
    &=2\iint\limits_D\mathrm{d}x\mathrm{d}y
\end{aligned}
$$
事实上,$D$的边界(消去$z$)为$x^2+y^2-2xy+4x+4y=4$.记$f(x,y)=x^2+y^2-2xy.$即将$\left[\begin{array}{cc}
    2 & -1 \\
    -1 & 2
\end{array}\right]$化为标准型:$|\lambda I-A| = 0$解得$\lambda_1=3,\lambda_2=1$.从而解得变换$T$:$\left[\begin{array}{cc}
    \dfrac{1}{\sqrt{2}} & \dfrac{1}{\sqrt{2}}\\
    -\dfrac{1}{\sqrt{2}} & \dfrac{1}{\sqrt{2}}
\end{array}\right]$,$\left[\begin{array}{c}
    x\\
    y
\end{array}\right]=T\left[\begin{array}{c}
    x^*\\
    y^*
\end{array}\right]$.从而有$f(x,y)+4(x+y)=3(x^*)^2+(y^*)^2+4\sqrt{2}y^*=4$,配方,平移坐标轴可得$3u^2+v^2=12$即$\dfrac{u^2}{4}+\dfrac{v^2}{12}=1,S=\iint\limits_\Omega\mathrm{d}u\mathrm{d}v=4\sqrt{3}\pi$.从而
$$
2\iint\limits_D\mathrm{d}x\mathrm{d}y=2\cdot 4\sqrt{3}\pi=8\sqrt{3}\pi
$$

\section{求球面 $x^{2}+y^{2}+z^{2}=2 R z$ 包含在锥面 $z^{2}=3\left(x^{2}+y^{2}\right)$ 内部分曲面的面积.}
\textbf{解}\quad
计算可得投影区域$D$为$x^2+y^2\leqslant \dfrac{3R^2}{4}$.其中的球面上每个点对$z$轴的夹角余弦为 \\$\dfrac{R}{\sqrt{R^2-x^2-y^2}}$.从而有
$$
\begin{aligned}
    \iint\limits_D\dfrac{1}{|\cos<\boldsymbol{n},z>|}\mathrm{d}x\mathrm{d}y&=\int_0^{2\pi}\mathrm{d}\theta\int_0^{\sqrt{3}R/2}\rho\dfrac{R}{\sqrt{R^2-\rho^2}}\mathrm{d}\rho\\
    &=\pi R^2
\end{aligned}
$$

\section{已知球体 $x^{2}+y^{2}+z^{2} \leqslant 2 R z$, 其上任意一点的密度等于该点到原点的距离的平方,求球体的质心坐标.}
\textbf{解}\quad
令$\eta = k(x^2+y^2+z^2)$,不妨令$k=1$.球坐标变换:$\left\{\begin{array}{l}
    x=\rho\sin\varphi\cos\theta\\
    y=\rho\sin\varphi\sin\theta\\
    z=\rho\cos\varphi
\end{array}\right.$,则$\dfrac{\partial(x,y,z)}{\partial(u,v,w)}=\rho^2\sin\varphi$.质心显然在$z$轴上,所以只考虑$z$轴上的分量即可.
$$
\begin{aligned}
    \dfrac{\iiint\limits_Dz\rho\mathrm{d}V}{\iiint\limits_D\rho\mathrm{d}V}&=\dfrac{\int_0^{2\pi}\mathrm{d}\theta\int_0^{\pi/2}\mathrm{d}\varphi\int_0^{2R\cos\theta}\rho^5\cos\varphi\sin\varphi\mathrm{d}\rho}{\int_0^{2\pi}\mathrm{d}\theta\int_0^{\pi/2}\mathrm{d}\varphi\int_0^{2R\cos\varphi}r^4\sin\varphi\mathrm{d}r}\\
    &=\dfrac{5R}{4}
\end{aligned}
$$
即质心在$(0,0,\dfrac{5R}{4})$处.
\section{求密度均匀的上半椭球体 $\dfrac{x^{2}}{a^{2}}+\dfrac{y^{2}}{b^{2}}+\dfrac{z^{2}}{c^{2}} \leqslant 1(z \geqslant 0)$ 的质心.}
显然质心在$z$轴上.
$$
\begin{aligned}
    \dfrac{\iiint\limits_\Omega z\mathrm{d}V}{V}&=\dfrac{\iiint\limits_\Omega\mathrm{d}x\mathrm{d}y\mathrm{d}z}{V}\\
    &=\dfrac{\int_0^{2\pi}\mathrm{d}\theta\int_0^{\pi/2}\mathrm{d}\varphi\int_0^1abc^2\rho^3\cos\varphi\sin\varphi\mathrm{d}\rho}{\dfrac{2}{3}\pi abc}\\
    &=\dfrac{3c}{8}
\end{aligned}
$$
即质心在$(0,0,\dfrac{3c}{8})$处.

\section{求质量为 $m$ 密度均匀的圆环 $D: r^{2} \leqslant x^{2}+y^{2} \leqslant R^{2}(r, R>0)$ 对垂直于圆环面中心轴的转动惯量.}
\textbf{解}\quad
\begin{tikzpicture}
    \filldraw(0,0)circle(2);
    \filldraw[white](0,0)circle(1);
    \draw[->](-0.2,0)--(3,0)node[right]{$x$};
    \draw[->](0,-0.2)--(0,3)node[above]{$y$};
\end{tikzpicture}

$\eta = \dfrac{m}{S}=\dfrac{m}{\pi(R^2-r^2)}$
$$
\begin{aligned}
    I=\iint\limits_D(x^2+y^2)\eta\mathrm{d}x\mathrm{d}y&=\eta\int_0^{2\pi}\mathrm{d}\theta\int_r^R\rho^3\mathrm{d}\rho\\
    &=\dfrac{m(R^2+r^2)}{2}
\end{aligned}
$$

\section{求质量为 $M$ 的均匀薄片 $\left\{\begin{array}{l}x^{2}+y^{2} \leqslant a^{2} \text { , } \\ z=0\end{array}\right.$ 对 $z$ 轴上的点 $(0,0, c)$ 处的单位质量质点的引力.}
\textbf{解}\quad
线密度:$\sigma=\dfrac{M}{\pi a^2}$
$$
\begin{aligned} 
    \iint\limits_D\dfrac{cG\sigma}{(x^2+y^2+c^2)^{3/2}}\mathrm{d}x\mathrm{d}y&=cG\sigma\int_0^{2\pi}\mathrm{d}\theta\int_0^{a}\dfrac{\rho}{(\rho^2+c^2)}\mathrm{d}\rho\\
    &=\dfrac{2cGM}{a^2}\left(\dfrac{1}{c}-\dfrac{1}{\sqrt{a^2+c^2}}\right)
\end{aligned}
$$



\end{document}